EXPLANATION OF ZIRCONIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 1, 2015 Zirconium is a chemical element with symbol Zr and atomic number 40. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Zirconium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d25s2. According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of zirconium (from (E1 to E5 ) are the following: E1 = 6.63, E2 = 13.13, E3 = 23, E4 = 34.34, and E5 = 80.35 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” published in Ind. J. Th. Phys. (2008). ' ' EXPLANATION OF ( E1 + Ε2 ) = 19.76 eV = -Ε(5s2) Here the E(5s2) represents the ground state energy of the two outermost electrons, (5s2), with opposite spin given by applying my formula of 2008. The charges (-38e) of the inner electrons (1s22s22p63s23p63d104s2 4p64d2 ) screen the nuclear charge (+40e) and for a perfect screening we would have ζ = 2. However the two electrons of 5s2 penetrate the 4d2 and 4p6 and lead to the deformations of spherical electron clouds. Thus ζ > 2. Under this condition we may write ( E1 + E2) = 19.76 eV = [ (27.21)ζ2 - (16.95)ζ + 4.1]/ n2 Since n = 5 this equation can be rewritten as 1.09 ζ2 - 0.678 ζ – 19.6 = 0 Τhen solving for ζ we get ζ = 4.56 > 2. Note that the two electrons of opposite spin (5s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. This situation occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. ' ' EXPLANATION OF ( Ε3 + E4 ) = 57.34 eV = - E(4d2) Here the E(4d2) represents the binding energy of the two electrons of parallel spin given by applying the Bohr formula. The charges (-36e) of (1s22s22p63s23p63d104s2 4p6) screen the nuclear charge (+40e) and for a perfect screening we would have ζ = 4. However the two electrons (4d2) repel the electrons of 4p6 and lead to the deformations of spherical electron clouds. Thus ζ > 4. Under this condition we write ( E3 + E4 ) = 57.34 eV = - E(4d2) = -2(-13.6057) ζ2/ n2 Since n = 4 we get ζ = 5.8 > 4 EXPLANATION OF Ε5 = 80.35 eV ' '= -E(4p6) + E(4p5) The charges (-30e) of (1s22s22p63s23p63d104s2 ) screen the nuclear charge (+40e) and for a perfect screening we would have ζ =10. However the electrons of 4p repel the electrons of 4s2 and lead to the deformations of spherical electron clouds. Thus ζ > 10. Note that 4p6 consists of three pairs (6 electrons). Thus we may write -E(4p6) = -3ζ2 + (16.95) ζ - 4.1 / n2 On the other hand, since 4p5 consists of two pairs (4 electrons ) and of one electron we write E(4p5) = 2+ (16.95)ζ - 4.1/n2 + (-13.6057)ζ2/n2 Therefore E5 = 80.35 eV = -E(4p6) + E(4p5) = (13.6ζ2 -16.95ζ + 4.1) / n2 Since n = 4 the above equation can be written as 0.85 ζ2 - 1.06 ζ - 80.1 = 0 Then solving for ζ we get ζ = 10.35 > 10 . Category:Fundamental physics concepts